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seropon [69]
3 years ago
7

The average speed of an object, S SS, is calculated using the formula S = D T S= T D ​ S, equals, start fraction, D, divided by,

T, end fraction, where T TT is the time it takes to travel a distance of D DD units. Rearrange the formula to solve for time ( T ) (T)
Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

T = D/S

Explanation:

Given the formula for calculating the average speed of an object as:

average speed = Distance/Time

Let average speed = S

Distance = D

Time = T

Substitute

S = D/T

Make T the subject of the formula:

From S = D/T

Cross multiply

ST = D

Divide both sides by S

ST/S = D/S

T = D/S

Hence the expression for he time T after rearrangement is T = D/S

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Find the magnitude of the impulse delivered to a soccer ball when a player kicks it with a force of 1360 N. Assume that the play
Alona [7]

Answer:

7.59Ns

Explanation:

Given parameters:

Force  = 1360N

Time of contact  = 5.85 x 10⁻³s

Unknown:

Impulse  = ?

Solution:

The impulse of the ball is given as:

        Impulse  = Force x time

       Impulse  = 1360 x 5.85 x 10⁻³ = 7.59Ns

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3 years ago
A team of scientists wants to conduct a study on an endangered animal inthe wild. Which rule should the study follow in order to
cricket20 [7]

Answer:

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3 years ago
Who was the first person to discover wormholes?
valentinak56 [21]

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8 0
3 years ago
Read 2 more answers
007 (part 1 of 2) 1.0 points
levacccp [35]

Answer:

a) The angle of refraction is approximately 34.7

b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°

Explanation:

According to Snell's law, we have;

\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}

The refractive index of the glass, n₁ = 1.66

The angle of incident of the light as it moves into water, θ₁ = 27.2°

a) The refractive index of water, n₂ = 1.333

Let θ₂ represent the angle of refraction of the light in water

By plugging in the values of the variables in Snell's Law equation gives;

\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}

sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265

θ₂ = arcsin(0.5692292265) ≈ 34.7°

The angle of refraction of the light in water, θ₂ ≈ 34.7°

b) When the angle of refraction, θ₂ = 90°, we have;

\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}

sin (\theta_1) = \dfrac{sin (90^{\circ})}{\left( \dfrac{1.66}{1.333}\right)} = sin (90^{\circ}) \times \dfrac{1.333}{1.66} \approx 0.803

θ₁ ≈ arcsin(0.803) ≈ 53.42°

The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°

3 0
3 years ago
You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?
Inga [223]

Answer:

The max height of the ball is y = -1/2 (v0²/g).

It takes the ball t = -2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = -v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let´s find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 =  v0 + g · t

t = -v0/g

Now, replacing t =  -v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²   (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g ·  (-v0/g)²

y = -(v0²/g) + 1/2 · (v0²/g)

y = -1/2 (v0²/g)

The max height of the ball is y = -1/2 (v0²/g).  Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = -2 v0/g.

However, let´s calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 =  v0 · t + 1/2 · g · t²

0 = t · ( v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = -2 · v0/g to hit the ground.

Let´s use the equation of velocity at final time (t = -2 · v0/g):

v = v0 + g · t

v = v0 + g · ( -2 · v0/g)

v = v0 - 2· v0

v = -v0

The speed of the ball when it hits the ground is v = -v0.

7 0
4 years ago
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